Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `y = (2x+4)/( x^2-3x-4)`?

Answer 1

so we have vertical asymptotes at `x = -1, 4`

for horizontal and slope asmptotes, `lim_{x \to pm oo} y = 0`

Explanation:

for vertical aympptotes, we look at when the demoninator is zero

so
`x^2 - 3x - 4 = (x+ 1)(x - 4) implies x = -1, 4`

to check for possible indeterminates we note that

`y(-1) = 2/0` = ndef
and
`y(4) = 12/0` = ndef

so we have vertical asymptotes at `x = -1, 4`

for horizontal and slope we look at the behaviour of the function as `x \to pm oo`

so we re-write
`lim_{x to pm oo} (2x+4)/(x^2 - 3x - 4)`

as

`lim_{x to pm oo} (2/x+4/x^2)/(1 - 3/x - 4/x^2) = 0`